$\gcd\left(\binom{n}{k},n\right)>1$ for nontrivial values? Is the following conjecture true?

Given an $n>1$ and a $k$ such that $0<k<n$ then $\gcd\left(\binom{n}{k}, n\right)>1$

This was a question asked in my old Number Theory course that I recently remembered. It was posed after we learned that $p\mid\binom{p}{k}$ for $0<k<p$. We never resolved it either way. I expect it to be true and have an elegant answer, but I haven't been able to prove it. 
 A: More generally: $$\gcd\left(\binom{n}{i}, \binom{n}{j}\right) > 1 \mbox{ if } 0 < i, j < n.$$
Proof: The following identity is easily verified: $$\binom{n}{i} \binom{n-i}{j} = \binom{n}{j} \binom{n-j}{i}.$$  Now if $\gcd\left(\binom{n}{i}, \binom{n}{j}\right) = 1$, then the above identity implies $\binom{n}{i} | \binom{n-j}{i}$ which is impossible because $\binom{n}{i} > \binom{n-j}{i}$.  Hence $\gcd\left(\binom{n}{i}, \binom{n}{j}\right) > 1$.
This argument comes from a paper of Erdős and Szekeres. (The identity they actually use is slightly different and they prove a stronger claim, that the gcd is $\ge 2^i$ for $i, j \le n/2$.)
A: Note that
$$
\begin{align}
\binom{n}{k}
&=\frac{n}{k}\binom{n-1}{k-1}\\
&=\frac{n}{\gcd(k,n)}\frac{\gcd(k,n)}{k}\binom{n-1}{k-1}\tag{1}
\end{align}
$$
Therefore, since $(1)$ is an integer,
$$
\left.\frac{k}{\gcd(k,n)}\,\middle|\,\frac{n}{\gcd(k,n)}\binom{n-1}{k-1}\right.\tag{2}
$$
However, since
$$
\gcd\left(\frac{n}{\gcd(k,n)},\frac{k}{\gcd(k,n)}\right)=1\tag{3}
$$
$(2)$ implies we must have
$$
\left.\frac{k}{\gcd(k,n)}\,\middle|\,\binom{n-1}{k-1}\right.\tag{4}
$$
Therefore, $(1)$ and $(4)$ show that
$$
\left.\frac{n}{\gcd(k,n)}\,\middle|\,\binom{n}{k}\right.\tag{5}
$$
and thus,
$$
\left.\frac{n}{\gcd(k,n)}\,\middle|\,\gcd\left(\binom{n}{k},n\right)\right.\tag{6}
$$
If $0\lt k\lt n$, then $\gcd(k,n)\lt n$ and therefore $\dfrac{n}{\gcd(k,n)}\gt1$.
A: See OEIS sequence A091963 and reference given there.
